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Unformatted text preview: Math 523H Homework #3 Sections 1.4, 2.1 1.4 #1. Give 5 examples which show that P implies Q does not necessarily mean that Q implies P. If x R and x > 0, then x 2 > 0. However, if x 2 > 0, x is not necessarily greater than 0. If x R and x 0, then  x  0. But if  x  0, then x is not necessarily greater than or equal to 0. If x and y are even integers, then x + y is an even integer. But if x + y is an even integer, then x and y are not necessarily even. If a and b are rational numbers, then a b is a rational number. But if a b is a rational number, then a and b are not necessarily rational. If x and y are both even integers, then xy is an even integer. But if xy is an even integer, then it is not necessarily true that both x and y are even. 1.4 #4. Suppose that < a < b . Prove that: (a) a < ab < b . Proof. a < b . Thus, a a < a b , or a 2 < ab . Since a > 0 and b > 0, ab > 0 so ab exists in R . Thus, we can are able to take the square root of both sides, and we get a 2 < ab , or a < ab . Since a < b , we also have a b < b b , or ab < b 2 . So ab < b 2 , or ab < b . By transitivity, a < ab < b . (b) ab 1 2 ( a + b ). Proof. For any a,b R , 0 ( a b ) 2 , or 0 a 2 2 ab + b 2 . By (O4) of section 1.1, 0 + 4 ab ( a 2 2 ab + b 2 ) + 4 ab , or 4 ab a 2 + 2 ab + b 2 . But a 2 + 2 ab + b 2 = ( a + b ) 2 , so 4 ab ( a + b ) 2 . Since a > 0 and b > 0, 4 ab > 0 and thus 4 ab exists in R . Thus, we can take the square root of both sides, and 4 ab p ( a + b ) 2 . This may be simplified to 2 ab < a + b . Since 1 2 > 0, we can use (O5) of section 1.1, and we get 1 2 (2 ab ) 1 2 ( a + b ), or ab 1 2 ( a + b ). 1.4 #7. Prove that for any n natural numbers, x 1 ,x 2 ,...,x n ,  x 1 + x 2 + ... + x n   x 1  +  x 2  + ... +  x n  . Proof. For n = 1,  x 1  =  x 1  , so the inequality holds for n = 1. Assume the above statement is also true for the first n natural numbers.  x 1 + x 2 + ... + x n + x n +1  =  ( x 1 + x 2 + ... + x n )+ x n +1  , so by the Triangle Inequality,  ( x 1 + x 2 + ... + x n )+ x n +1   x 1 + x 2 + ... + x n  +  x n +1  . Because we assumed that the statement above was true for the first n natural numbers,  x 1 + x 2 + ... + x n   x 1  +  x 2  + ... +  x n  . Therefore,  x 1 + x 2 + ... + x n  +  x n +1   x 1  +  x 2  + ... +  x n  +  x n +1  . By transitivity,  x 1 + x 2 + ... + x n + x n +1   x 1  +  x 2  + ... +  x n  +  x n +1  , so the statement is also true for the first n +1 natural numbers. By induction, we have proven that  x 1 + x 2 + ... + x n   x 1  +  x 2  + ... +  x n  for any n N ....
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This note was uploaded on 12/07/2010 for the course MATH 523 taught by Professor Nahmod during the Fall '10 term at UMass (Amherst).
 Fall '10
 Nahmod
 Math

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